The Bolzano–Weierstrass theorem states that every bounded real sequence contains at least one convergent subsequence.
The theorem captures a deep property of the real line: a sequence confined to a closed, bounded interval cannot spread out without bound. Even when the sequence itself fails to converge, one can always extract a subsequence that does.
Contents
- Statement of the Bolzano–Weierstrass theorem
- What the theorem means
- Proof by nested intervals
- Why the boundedness hypothesis is needed
- Worked examples
- Connection with accumulation points
Statement of the Bolzano–Weierstrass theorem
Consider a real sequence
\[ (x_n)_{n\in\mathbb N}. \]
Recall that a sequence is said to be bounded if there exist two real numbers \(a\) and \(b\), with \(a\leq b\), such that
\[ a\leq x_n\leq b \]
for every \(n\in\mathbb N\). In other words, all the terms of the sequence lie within a single closed, bounded interval \([a,b]\).
Bolzano–Weierstrass theorem. Every bounded real sequence admits a convergent subsequence.
Equivalently, if \((x_n)\) is a bounded real sequence, then there exist a strictly increasing sequence of indices
\[ n_1\lt n_2\lt n_3\lt\cdots \]
and a real number \(x_0\) such that
\[ x_{n_k}\longrightarrow x_0. \]
The sequence \((x_{n_k})\) is called a subsequence of \((x_n)\).
What the theorem means
The theorem does not claim that every bounded sequence converges — that would be false. Take, for instance, the sequence
\[ x_n=(-1)^n, \]
which is bounded but does not converge, since it oscillates endlessly between \(1\) and \(-1\).
It does, however, contain convergent subsequences. Indeed, taking the even-numbered indices gives
\[ x_{2k}=1 \]
for every \(k\), and therefore
\[ x_{2k}\longrightarrow 1. \]
Taking the odd-numbered indices instead gives
\[ x_{2k-1}=-1 \]
for every \(k\), and therefore
\[ x_{2k-1}\longrightarrow -1. \]
This is precisely what the Bolzano–Weierstrass theorem asserts: even when a bounded sequence fails to converge as a whole, somewhere inside it there is always a subsequence that does.
Proof by nested intervals
We prove the theorem using the Nested Interval Theorem.
Let \((x_n)\) be a bounded real sequence. Then there exist \(a,b\in\mathbb R\), with \(a\leq b\), such that
\[ x_n\in[a,b] \]
for every \(n\in\mathbb N\).
Set
\[ I_1=[a,b]. \]
The interval \(I_1\) contains every term of the sequence, so it certainly contains infinitely many of them.
Split \(I_1\) into two closed intervals of equal length:
\[ \left[a,\frac{a+b}{2}\right], \qquad \left[\frac{a+b}{2},b\right]. \]
Since \(I_1\) contains infinitely many terms of the sequence, at least one of the two subintervals must contain infinitely many of them as well. We choose such a subinterval and call it \(I_2\).
We now repeat the procedure. Suppose we have built a closed interval \(I_k\) containing infinitely many terms of the sequence. We split \(I_k\) into two closed intervals of equal length. At least one of them contains infinitely many terms; we select it and call it \(I_{k+1}\).
In this way we obtain a sequence of closed, bounded intervals
\[ I_1\supseteq I_2\supseteq I_3\supseteq\cdots \]
in which each \(I_k\) contains infinitely many terms of the sequence \((x_n)\).
Moreover, at each step the length of the interval is halved. If \(I_1=[a,b]\), then the length of \(I_k\) is
\[ \frac{b-a}{2^{k-1}}. \]
Since
\[ \frac{b-a}{2^{k-1}}\longrightarrow 0, \]
the Nested Interval Theorem guarantees a unique point \(x_0\in\mathbb R\) such that
\[ \bigcap_{k=1}^{+\infty} I_k=\{x_0\}. \]
It remains to construct a subsequence of \((x_n)\) that converges to \(x_0\).
We build this subsequence by induction on the intervals \(I_k\).
Since \(I_1\) contains infinitely many terms of the sequence, we choose an index \(n_1\) with
\[ x_{n_1}\in I_1. \]
Since \(I_2\) contains infinitely many terms, we can choose an index \(n_2\) greater than \(n_1\) with
\[ x_{n_2}\in I_2. \]
In general, suppose we have already chosen indices
\[ n_1\lt n_2\lt\cdots\lt n_k \]
so that
\[ x_{n_j}\in I_j \qquad \text{for every } j=1,\ldots,k. \]
Since \(I_{k+1}\) contains infinitely many terms, we can pick an index \(n_{k+1}\gt n_k\) with
\[ x_{n_{k+1}}\in I_{k+1}. \]
We thereby obtain a subsequence
\[ (x_{n_k})_{k\in\mathbb N} \]
such that
\[ x_{n_k}\in I_k \qquad \forall k\in\mathbb N. \]
We now show that this subsequence converges to \(x_0\).
Since \(x_0\) belongs to every interval \(I_k\) and \(x_{n_k}\in I_k\) as well, the distance between \(x_{n_k}\) and \(x_0\) is at most the length of \(I_k\). Hence
\[ |x_{n_k}-x_0|\leq \frac{b-a}{2^{k-1}}. \]
Since
\[ \frac{b-a}{2^{k-1}}\longrightarrow0, \]
the squeeze theorem yields
\[ |x_{n_k}-x_0|\longrightarrow0. \]
Therefore
\[ x_{n_k}\longrightarrow x_0. \]
We have thus exhibited a convergent subsequence of the original sequence, which completes the proof of the Bolzano–Weierstrass theorem.
Why the boundedness hypothesis is needed
The boundedness hypothesis is essential. If a sequence is unbounded, it need not admit any convergent subsequence.
Consider, for example, the sequence
\[ x_n=n. \]
It is not bounded above. Furthermore, every one of its subsequences has the form
\[ x_{n_k}=n_k, \]
where
\[ n_1\lt n_2\lt n_3\lt\cdots. \]
Since the indices \(n_k\) tend to \(+\infty\), we have
\[ x_{n_k}=n_k\longrightarrow+\infty. \]
No subsequence can therefore converge to a real number.
This example shows that boundedness is no mere technicality: it is exactly what keeps the terms of the sequence from escaping to infinity.
Worked examples
Example 1. Consider the sequence
\[ x_n=(-1)^n. \]
The sequence is bounded, since
\[ -1\leq x_n\leq1 \]
for every \(n\in\mathbb N\). By the Bolzano–Weierstrass theorem, it admits at least one convergent subsequence.
Indeed, the even-indexed terms give
\[ x_{2k}=1 \]
for every \(k\in\mathbb N\), so that
\[ x_{2k}\longrightarrow1, \]
while the odd-indexed terms give
\[ x_{2k-1}=-1, \]
and therefore
\[ x_{2k-1}\longrightarrow -1. \]
The original sequence does not converge, yet it has two natural convergent subsequences.
Example 2. Consider the sequence
\[ x_n=\frac{(-1)^n n}{n+1}. \]
It is bounded, because
\[ -1\lt x_n\lt1 \]
for every \(n\in\mathbb N\). By Bolzano–Weierstrass, it must admit a convergent subsequence.
We separate the even and odd indices. If \(n=2k\), then
\[ x_{2k}=\frac{2k}{2k+1}\longrightarrow1. \]
If instead \(n=2k-1\), then
\[ x_{2k-1}=-\frac{2k-1}{2k}\longrightarrow -1. \]
Here too the sequence fails to converge, but it does contain convergent subsequences.
Example 3. Consider an arbitrary sequence \((x_n)\) contained in the interval \([0,1]\).
There is no need to know an explicit formula for the sequence. The mere fact that
\[ 0\leq x_n\leq1 \]
for every \(n\in\mathbb N\) guarantees, by the Bolzano–Weierstrass theorem, the existence of a convergent subsequence.
This is one of the most important features of the theorem: it provides an existence result even when we have no way to compute a subsequence explicitly.
Connection with accumulation points
The Bolzano–Weierstrass theorem can also be read in terms of accumulation points.
If a bounded real sequence takes infinitely many distinct values, then its set of values is an infinite, bounded subset of \(\mathbb R\). In that case the theorem guarantees the existence of at least one accumulation point.
More precisely, if a subsequence
\[ x_{n_k}\longrightarrow x_0 \]
and the terms \(x_{n_k}\) differ from \(x_0\) for infinitely many indices, then \(x_0\) is an accumulation point of the set of values of the sequence.
If instead the sequence takes only finitely many values, the theorem still holds: at least one of those values must be attained infinitely often. In that case there is a constant — and hence convergent — subsequence.
Bolzano–Weierstrass can thus be understood in two complementary ways:
- every bounded real sequence has a convergent subsequence;
- every infinite, bounded set of real numbers has at least one accumulation point.
This second formulation links the theorem to the topological study of the real line and sets the stage for results on compactness.